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Let $E$ is a norm space. $\varnothing \neq A \subset E$ is open,convex set. $M$ is subspace of $E$. Suppose $A \cap M = \varnothing$. Then there exist a closed hyperplane $H$ such that $M \subset H$ and $H \cap M= \varnothing$

I see Hahn-Banach first geometric form: "Let $A \subset E$ and $B \subset E$ be nonempty convex subsets such that $A \cap B= \varnothing$. Assume that one of them is open. Then exist a closed hyperplane that separates $A$ and $B$".

Is it used to prove above proposition?

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According to the Hahn-Banach first geometric form, $\exists l\in E^\ast\setminus\{0\}$, $c\in\mathbb{R}$ s.t. \begin{eqnarray} \forall x\in A, y\in M \quad l(x)\leq c \leq l(y). \end{eqnarray} Then we have $l(M)=\{l(0)\}$ from the second inequality above. Thus putting $H:=l^{-1}(l(0))$, we have done.

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  • $\begingroup$ Thank you but why $l(M)=\{l(0) \}$? $\endgroup$ – Nguyễn Đăng Khải Hoàn Apr 30 '17 at 9:02
  • $\begingroup$ Assume that $\exists y_0, y_1 \in M$ s.t. $l(y_0)<l(y_1)$ for a while. Then $l(t(y_0-y_1))$ goes to $-\infty$ as $t\to\infty$ while $t(y_0-y_1)\in M$. This contradicts to the latter inequality in my answer. $\endgroup$ – stb2084 Apr 30 '17 at 9:11
  • $\begingroup$ that's right. Thank you stb2084 $\endgroup$ – Nguyễn Đăng Khải Hoàn Apr 30 '17 at 9:40

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